Topology/Vector spaces

"Let V and W be vector spaces and T: V—>W be a linear transformation. Then,

T is one-to-one if and only if T carries linearly independent subsets of V onto linearly independent subsets of W"

My question:
Let {w1} be a linearly independent subset of W which does not belong to R(T) [i.e., range of T]. Therefore, no vector v exists in V such that T(v)=w. So the theorem's claim that T carries every L.I. subset of V "onto" every L.I. subset of "W" send incorrect.

The theorem would make sense if it either used "into" instead of "onto" or if it used "R(T)" instead of "W".

It would be great if you could help be in resolving this doubt!

Thanks and regards.

ANSWER: It seems like you have misunderstood the premise. It only says that T sends independent vectors in V to independent vectors in W . So , for example, if v1 , v2 are independent in V , then T(v1) and T(v2) are independent in W . It does not say that if w1 and w2 are independent vectors in W , then they must be the images of some independent vectors in V . As you correctly point out, w1 and w2 may not even be in T(V)

---------- FOLLOW-UP ----------

QUESTION: Hi Socrates,

My contention is with the word "onto". It says that 'T carries L.I. subsets of V "onto" L.I. subsets of W'

Based on the definition of the word "onto" used in the context of maps/functions/transformations, I inferred that the theorem implies that every L.I. Subset of W must have a primate in V.

AnswerFor your interpretation, it would be stated "T carries L.I.subsets of V onto all L.I. subsets of W"

Or another way

T carries the collection of L.I. subsets
of V onto the collection of L.I subsets of W

Expertise

I can answer most questions on Point Set Topology, but not in Algebraic Topology. Questions of the sort given as exercises in any of the standard Topology texts or Ph.D. "prelim" level problems are welcome.

Experience

My Ph.D. thesis was in Algebra, but I had three graduate courses in Point Set Topology which I enjoyed very much. I am also an experienced teacher.